# Metric tensor with components

1. Sep 7, 2014

### ChrisVer

I have one question, which I don't know if I should post here again, but I found it in GR...
When you have a metric tensor with components:
$g_{\mu \nu} = \eta _{\mu \nu} + h_{\mu \nu}, ~~ |h|<<1$ (weak field approximation).

Then the inverse is:

$g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$ right? However that doesn't give exactly that $g^{\mu \rho}g_{\rho \nu} = \delta^{\mu}_{\nu}$ because of the existence of the $- h^{\mu \rho}h_{\rho \nu}$ which is of course small but it's not zero... Can the inverse matrix be defined approximately?

Also I don't understand why should the derivatives of $h$ behave as $h$ itself? I mean they take the terms like $h \partial h, ~~ \partial h \partial h$ to be of order $\mathcal{O}(h^2)$... why?

Last edited: Sep 7, 2014
2. Sep 7, 2014

### samalkhaiat

They mean to write
$$g_{ \mu \nu } = \eta_{ \mu \nu } + \epsilon h_{ \mu \nu } , \ \ \ |\epsilon | \ll 1.$$
Then all the following holds
$$g \cdot g = \delta + \mathcal{O}( \epsilon^{2} ) , \ \ h \cdot h \sim h \partial h \sim \partial h \cdot \partial h \sim \mathcal{O} ( \epsilon^{ 2 } ) .$$